/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q102SE An educational consulting firm i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Q102SE (page 137)

An educational consulting firm is trying to decide whether high school students who have never before used a hand-held calculator can solve a certain type of problem more easily with a calculator that uses reverse Polish logic or one that does not use this logic. A sample of \({\rm{25}}\) students is selected and allowed to practice on both calculators. Then each student is asked to work one problem on the reverse Polish calculator and a similar problem on the other. Let \({\rm{p = P(S)}}\), where \({\rm{S}}\) indicates that a student worked the problem more quickly using reverse Polish logic than without, and let \({\rm{X = }}\)number of \({{\rm{S}}^{\rm{'}}}{\rm{s}}\).

a. If \({\rm{p = }}{\rm{.5}}\), what is \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)?

b. If \({\rm{p = }}{\rm{.8}}\), what is \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)?

c. If the claim that \({\rm{p = }}{\rm{.5}}\) is to be rejected when either \({\rm{x}} \le {\rm{7}}\) or \({\rm{x}} \ge {\rm{18}}\), what is the probability of rejecting the claim when it is actually correct?

d. If the decision to reject the claim \({\rm{p = }}{\rm{.5}}\) is made as in part (c), what is the probability that the claim is not rejected when \({\rm{p = }}{\rm{.6}}\)? When \({\rm{p = }}{\rm{.8}}\)?

e. What decision rule would you choose for rejecting the claim \({\rm{p = }}{\rm{.5}}\) if you wanted the probability in part (c) to be at most \({\rm{.01}}\)?

Short Answer

Expert verified

(a) If\({\rm{p = }}{\rm{.5}}\), the value of\({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)is\({\rm{0}}{\rm{.986}}\).

(b) If\({\rm{p = }}{\rm{.8}}\), the value of\({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)is\({\rm{0}}{\rm{.220}}\).

(c) The probability of rejecting the claim \({\rm{p = }}{\rm{.5}}\) when it is actually correct is \({\rm{P(Rc) = 0}}{\rm{.044}}\)

(d) The probability of rejecting the claim\({\rm{p = }}{\rm{.6}}\)and\({\rm{p = }}{\rm{.8}}\)when they are actually correct is\({\rm{P(Rc) = 0}}{\rm{.845}}\)and\({\rm{P(Rc) = 0}}{\rm{.109}}\)respectively.

(e) The decision rule for rejecting the claim is\({\rm{X}} \le {\rm{5}}\)and\({\rm{X}} \ge {\rm{19}}\).

Step by step solution

01

Concept Introduction

The Binomial Random Variable\({\rm{X}}\)is defined as

\({\rm{X = }}\)the number of a month\({\rm{n}}\)trials

Where conditions one to four on page\({\rm{117}}\)are satisfied (binomial experiment).

The described random variable has Binomial Distribution with parameters

\({\rm{n = 25}}\)and\({\rm{p = P(S)}}\)

02

 Step 2: Calculation for \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)

(a)

The second parameter is given 鈥�

\({\rm{p = 0}}{\rm{.5}}\)

Which means that 鈥�

\({\rm{X}} \sim {\rm{Bin(25,0}}{\rm{.5)}}\)

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters\({\rm{n}}\)and\({\rm{p}}\)is 鈥�

\({\rm{B(x;n,p) = P(X}} \le {\rm{x) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\;\;\;{\rm{x = 0,1, \ldots ,n}}\)

The theorem is 鈥�

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Now compute the requested probability as 鈥�

\(\begin{aligned} P(7 \leqslant X \leqslant 18) &= B(18;25,0.5) - B(6;25,0.5) \\ & = 0.993 - 0.007 \\ &= 0.986 \\ \end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.986}}\).

03

Calculation for \({\rm{P(7}} \le {\rm{X}} \le {\rm{18)}}\)

(b)

The second parameter is given 鈥�

\({\rm{p = 0}}{\rm{.8}}\)

Which means that 鈥�

\({\rm{X}} \sim {\rm{Bin(25,0}}{\rm{.8)}}\)

Cumulative Density Function cdf of binomial random variable\({\rm{X}}\)with parameters\({\rm{n}}\)and\({\rm{p}}\)is 鈥�

\({\rm{B(x;n,p) = P(X}} \le {\rm{x) = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\;\;\;{\rm{x = 0,1, \ldots ,n}}\)

The theorem is 鈥�

\({\rm{b(x;n,p) = }}\left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{{\rm{(1 - p)}}}^{{\rm{n - x}}}}}&{{\rm{,x = 0,1,2, \ldots ,n}}}\\{\rm{0}}&{{\rm{, otherwise }}}\end{array}} \right.\)

Now compute the requested probability as 鈥�

\(\begin{aligned}P(7 \leqslant X \leqslant 18) &= B(18;25,0.8) - B(6;25,0.8) \\ &= 0.220 - 0.000 \\ &= 0.220 \\ \end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.220}}\).

04

Probability of Rejecting the Claim

(c)

The claim is rejected if 鈥�

\(\begin{array}{c}{\rm{X}} \le {\rm{7}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

It is actually correct, so we use probability \({\rm{p = 0}}{\rm{.5}}\) to compute the probability of event 鈥�

\({\rm{Rc = }}\){rejecting the claim}

Thus, it is obtained 鈥�

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 7\} \cap \{ X \geqslant 18\} ) = P(X \leqslant 7) + P(X \geqslant 18) \\ &= P(X \leqslant 7) + [1 - P(X \leqslant 17)] \\ &= B(7;25,0.5) + 1 - B(17;25,0.5) \\ &= 0.022 + 1 - 0.978 \\ &= 0.044 \\ \end{aligned} \)

Therefore, the value is obtained as \({\rm{0}}{\rm{.044}}\).

05

Probability of Rejecting the Claim

(d)

Complement of event\({\rm{Rc}}\)is event 鈥�

\({\rm{8}} \le {\rm{X}} \le {\rm{17}}\)

When\({\rm{p = 0}}{\rm{.6}}\), the probability that the claim is not rejected is 鈥�

\(\begin{aligned} P(8 \leqslant X \leqslant 17) &= B(17;25,0.6) - B(7;25,0.6) \\ &= 0.846 - 0.001 \\ &= 0.845 \\ \end{aligned} \)

When\({\rm{p = 0}}{\rm{.8}}\), the probability that the claim is not rejected is 鈥�

\(\begin{aligned}P(8 \leqslant X \leqslant 17) &= B(17;25,0.8) - B(7;25,0.8) \\ &= 0.109 - 0.000 \\ & = 0.109 \\ \end{aligned} \)

Therefore, the values are obtained as \({\rm{0}}{\rm{.845}}\) and \({\rm{0}}{\rm{.109}}\).

06

The Decision Rule

(e)

From the relation it is obtained 鈥�

\({\rm{P(Rc)}} \le {\rm{0}}{\rm{.01}}\)

Find values of random variable\({\rm{X}}\)that would yield such probability.

When the decision rule is 鈥�

\(\begin{array}{c}{\rm{X}} \le {\rm{7}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

Probability of\({\rm{0}}{\rm{.044}}\)is obtained, therefore smaller decision rule is needed, for example 鈥�

\(\begin{array}{c}{\rm{X}} \le {\rm{6}}\\{\rm{X}} \ge {\rm{18}}\end{array}\)

Using this, the following probability is obtained 鈥�

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 6\} \cap \{ X \geqslant 18\} ) = P(X \leqslant 6) + P(X \geqslant 18) \\ &= P(X \leqslant 6) + [1 - P(X \leqslant 17)] \\ & = B(6;25,0.5) + 1 - B(17;25,0.5) \\ &= 0.007 + 1 - 0.978 \\ & = 0.029 \\\end{aligned} \)

Which is still not small enough. Then try another decision rule 鈥�

\(\begin{array}{c}{\rm{X}} \le {\rm{5}}\\{\rm{X}} \ge {\rm{19}}\end{array}\)

Using this, the following probability is obtained 鈥�

\(\begin{aligned} P(Rc) &= P(\{ X \leqslant 5\} \cap \{ X \geqslant 19\} ) = P(X \leqslant 5) + P(X \geqslant 19) \\ &= P(X \leqslant 5) + [1 - P(X \leqslant 18)] \\ &= B(5;25,0.5) + 1 - B(18;25,0.5) \\&= 0.002 + 1 - 0.993 \\ &= 0.009 \\ \end{aligned} \)

Therefore, the last decision rule can be used.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha = 10\)per hour. Suppose that with probability \(.{\bf{5}}\) an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed \(y \ge 10\), what is the probability that y arrives during the hour, of which ten have no violations? c. What is the probability that ten 鈥渘o-violation鈥� cars arrive during the next hour?

A k-out-of-n system is one that will function if and only if at least k of the n individual components in the system function. If individual components function independently of one another, each with probability.\(9\), what is the probability that a 3-out-of-5 system functions?

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives circuit boards in batches of five. Two boards are selected from each batch for inspection. We can represent possible outcomes

of the selection process by pairs. For example, the pair (1, 2) represents the selection of boards 1 and 2 for inspection.

a. List the ten different possible outcomes.

b. Suppose that boards 1 and 2 are the only defective boards in a batch. Two boards are to be chosen at random. Define X to be the number of defective boards observed among those inspected. Find the probability distribution of X.

c. Let F(x) denote the cdf of X. First determine \(F\left( 0 \right) = P\left( {X \le 0} \right)\), F(1), and F(2); then obtain F(x) for allother x.

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday鈥檚 mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one\(P\left( {Wed.} \right) = 0.3\), \(P\left( {Thurs.} \right) = 0.4\), \(P\left( {Fri.} \right) = 0.2\), and\(P\left( {Sat.} \right) = 0.1\). Let Y = the number of days beyond Wednesday that it takes for both magazines to arrive (so possible Y values are 0, 1, 2, or 3). Compute the pmf of Y. (Hint: There are 16 possible outcomes; \(Y\left( {W,W} \right) = {\bf{0}}\),\(Y\left( {F,Th} \right) = 2\), and so on.)

Let X have a Poisson distribution with parameter \({\rm{\mu }}\). Show that E(X) =\({\rm{\mu }}\) directly from the definition of expected value. (Hint: The first term in the sum equals \({\rm{0}}\), and then x can be cancelled. Now factor out \({\rm{\mu }}\) and show that what is left sums to \({\rm{1}}\).)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.